Answer:
Step-by-step explanation:
Given that,
Metal of mass
M = 0.065kg
Initial temperature of metal
θm = 210°C
The metal is drop into a beaker which contain liquid of mass
m = 0.377 kg
Initial temperature of water
θw = 26°C
The final mixture temperature is
θf = 28.14°C
Specific heat capacity of water
Cw = 4186 J/kg°C
Since the metal is hotter than the water, then the metal will lose heat, while the water will gain heat, we assume that no heat is loss by the beaker.
So,
Heat Loss = Heat gain
Now, heat loss by metal
H(loss) = M•Cm•∆θ
Where M is mass of meta
Cm is specific capacity of metal, which we are looking fro
So,
H(loss) = 0.065 × Cm × (θi - θf)
H(loss) = 0.065 × Cm × (210-28.14)
H(loss) = 11.821 •Cm
Now, Heat gain by water
H(gain) = m•Cw•∆θ
H(gain) = m•Cw•(θf - θi)
Where
m is mass of water and Cw is specific heat capacity of water
H(gain) = 0.377 ×4286 × (28.14-26)
H(gain) = 3457.86
So, H(loss) = Heat(gain)
11.821 •Cm = 3457.86
Cm = 3457.86/11.821
Cm = 292.52 J/Kg°C
The specific heat capacity of the metal ball is 292.52 J/Kg°C